How do I solve y''+4y'+13y=3𝛿(t-π); y(0) =3, y'(0) =3?

Question

How do I solve y''+4y'+13y=3𝛿(t-π); y(0) =3, y'(0) =3? I am trying to solve for y(t).

Answer 1

We applying Laplace Transform: s²F(s) - 3s - 3 + 4(sF(s) - 3) + 13F(s) = 3e^-πs;

F(s) = (3s + 15 + 3e^-πs)/(s² + 4s + 13); F(s) = (3s + 15)/([(s + 2)² + 9] + e^-πs/[(s + 2)² + 9] = 3(s + 2)/[(s + 2)² + 3²] + 3·3/[(s + 2)² + 9] + ·3e^-πs/ [(s + 2)² + 9]. Now inverse Laplace gives solution:

y = f(t) = 3e^-2tcos3t + 3e^-2tsin3t + 1/3 + u(t - π)e^-πtsin3(t - π) = 3e^-2t(cos3t + sin3t) - u(t - π)e^-πtsin3t